Suppose we sample a function $f(x)$, but our resolution possesses a finite width $\Delta$: For each point $x_0$ we do not measure $f(x_0)$, but we obtain the moving average of the function $f$ with width $\Delta$. If we generalise this idea, and allow a second function $w(x)$ to describe the finite resolution, the moving average is identical to the convolution $$ g(x_0) = \int f(x-x_0) w(x) dx $$ If we use a rectangular function to describe the finite resolution, we obtain the initially described moving average, where each point within a certain distance to $x_0$ gets the same weight. However, if we like to a continuous function to describe the contribution to each point, the Gaussian is a "natural" choice, because it possesses the largest entropy.

Suppose we sample a function $f(x)$, but our resolution possesses a finite width $w=\Delta=const$: For each point $x_0$ we do not measure $f(x_0)$, but we obtain the moving average of the function $f$ with width $w$.

Next, we generalise this idea, and allow the finite resolution to be described by a function $w(x)$. We can think of it as a weighting function. At each point the same logic as above applies. However, instead of cutting out a finite part around $x=x_0$, and giving each point in this interval the same weight, the weight now depends of the distance to $x_0$. Thus, what we obtain is $$ g(x_0) = \int f(x-x_0) w(x) dx $$ which is identical to the convolution. Note, if we use a rectangular function to describe the finite resolution, we obtain the initially described moving average, where each point within a certain distance to $x_0$ gets the same weight. However, if we like to a continuous function to describe the contribution to each point, the Gaussian is a "natural" choice, because it possesses the largest entropy.